Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note: All numbers (including target) will be positive integers. Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak). The solution set must not contain duplicate combinations. For example, given candidate set 2,3,6,7 and target 7, A solution set is: [7] [2, 2, 3]
这道题看过好几次最开始都没看出怎么做就一直留着，结果今天在做Subsets 这道题时找到了灵感，使用深度优先的方法搜索，并且用一个vector记录向量，找到合适的向量时即将它保存在结果中，并进行回溯操作。这道题相比于Subsets中使用回溯法而言更麻烦一些。以后需要注意这种解题方法，当要求解的结果是一系列向量的集合时使用dfs搜索记录路径这种方法。对于输入candidates=[1,2] ，target=3，遍历的方向如图：
runtime:28ms
class Solution {public:vector<vector<int>> combinationSum(vector<int>& candidates, int target) {vector<vector<int>> result;vector<int> path;sort(candidates.begin(),candidates.end());helper(candidates,0,0,target,path,result);return result;}void helper(vector<int> &nums,int pos,int base,int target,vector<int>& path,vector<vector<int>> & result){if(base==target){result.push_back(path);return ;}if(base>target)return ;for(int i=pos;i<nums.size();i++){path.push_back(nums[i]);helper(nums,i,base+nums[i],target,path,result);path.pop_back();}}};
，同生天地间，为何我不能。
